Solution

WORK SHEET_LONG QUESTION

Class 12 - Physics
Section A
1. a.

b.

In right triangle ABC

BC
sin i =
AC

In △AEC sin r =
AE

AC

sin i BC v1 τ v1
= = = = μ
sin r AE v2 τ v2

c. Speed of yellow light inside the glass slab

c
v =
μ

8
3×10
= m/s
1.5
8
= 2 × 10 m/s

Wavelength of yellow light inside the glass slab

′ λ
λ =
μ

590
= nm
1.5

= 393.33 nm
2. Let PP' represents the surface separating medium 1 and medium 2 as shown in figure.

Let v1 and v2 represents the speed of light in medium 1 and medium 2 respectively. We assume a plane wavefront AB propagating
in the direction A'A incident on the interface at an angle of incidence i. Let t be the time taken by the wavefront to travel the

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 distance BC in medium 1 and AE in medium 2 respectively.
∴ BC = speed of light in medium 1 × time = v t [∵ distance = speed × time] 1

In order to determine the shape of the refracted wavefront, we draw an arc of radius v t from the point A in the second medium
2

(the speed of the wave in second medium is v and applying the formula, distance = speed × time). Let CE represents a tangent
2

plane drawn from the point C.

Then AE = speed of light in medium 2 × time = v t 2

∴ CE would represent the refracted wavefront.

In ΔABC and ΔAEC , we have

sin i v1 t AC sin i v1
= ⋅ ; ∴ =
sin r AC v2 t sin r v2

If c represents the speed of light in vacuum, then

μ =
1 and μ =
c

v1
[since the refractive index of a medium = speed of light in vacuum or air ÷ speed of light at that medium]
2
c

v2

⇒ v1 =
c

μ1
and v 2 =
c

μ2

where, µ1 and µ2 are the refractive indices of medium 1 and medium 2 respectively.
sin i c/μ1
∴ =
sin r c/μ2

sin i μ2
⇒ = ⇒ μ1 sin i = μ2 sin r
sin r μ
1

This proves the snell's law of refraction.

3. Suppose S1 and S2 are two fine slits, a small distance d apart. They are illuminated by a strong source S of monochromatic light of
wavelength λ . MN is a screen at a distance D from the slits.

Consider a point P at a distance y from 0, the centre of the screen.

The path difference between two waves arriving at point Pis equal to S2P - S1P.

Now, (S2P)2 -(S1P)2

2 2
2 d 2 d
= [D + (y + ) ] − [D + (y − ) ] = 2yd
2 2

2yd
Thus, S 2P − S1 P =
S2 P + S1 P

But S 2P + S1 P ≈ 2D
dy
∴ S2 P − S1 P ≈
D

a. For constructive interference (Bright fringes)

dy
Path difference = D
= nλ , where,
n = 0, 1, 2, 3,......
nDλ
∴ y =
d
[∵ n = 0, 1, 2, 3, ....]
b. For destructive interference (Dark fringes)
dy
Path difference = D
= (2n − 1)
λ

The distribution of intensity in Young's double slit experiment is as shown below

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4. i. Path difference in Young's Double slit experiment at point P is given by the equation:
Δx = S2 P − S1 P

2 2

S2P2 - S1P2 = [D 2
+ (x +
d

2
) ] − [D
2
+ (x −
d

2
) ]

= 2xd
(S2P - S1P)(S2P + S1P) = 2xd
Assuming S2P + S1P ≈ 2D as x << D and d << D
xd
Δx ≈
D

For constructive interference condition is, Δx = nλ

Position of nth bright fringe is given by: xn = nλD

and for destructive interference is given by, Δx = (2n + 1) λ

2
(2n+1)λD
Position of nth dark fringe is given by: xn = 2d

ii. Let intensity of light sources from slits be I.

Resultant intensity at a point is given by I' = I + I + 2Icos ϕ
where ϕ is the phase difference at the point. Path difference is given by:
λϕ
Δx =
2π

Hence, I' = I + I + 2Icos (2π Δx

λ
)

Given intensity at central maximum is Io

Hence, Io = 4I
λ 3
At Δx = 6
,I
′
= 3I =
4
Io

λ 1
At Δx= 4
,I ′
= 2I =
2
I0

At Δx = λ

3
,I
′
= I =
1

4
Io

5. The light on passing through the narrow slit undergoes diffraction. A diffraction pattern consisting of alternate bright and dark
bands is obtained on the screen.
i. Angular width of principal maximum,
2θ = 2λ

It is not affected when screen is moved away (D increases) from the slit plane.
ii. Now linear width x of the central maximum is given by
x= 2λD

Thus if the screen is moved away the linear width of the central maximum will increase too.
Difference between interference and diffraction
i. In interference all the fringes will be of equal intensity but in diffraction the central maximum will have high intensity and in
the rest of the fringes intensity falls rapidly.
ii. In interference all the fringes will be of equal width but in diffraction the central maximum will have the highest width and for
the other fringes width will diminish fast.

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6.

Initial path difference between S1 and S2,

Δ0 = SS2 - SS1 = λ

Path difference between disturbances from S1 and S2 at point P,

xd
Δ=
D

Total path difference between the two disturbances at P,

λ xd
ΔT = Δ0 + Δ = +
4 D

∴ For constructive interference,

λ xd
ΔT = (
4
+
D
) = nλ ; n = 0, 1, 2, ...
xn d
or D
= (n −
1

4
)λ ... (1)
For destructive interference,
ΔT = (
λ

4
+
xd

D
) = (2n - 1) λ

2
...(2)
′
xn d
or D
= (2n − 1 −
1

2
)
λ

2
′
xn d 3 λ
or D
= (2n −
2
)
2

λD
Fringe width, β = xn+1 - xn = d

The position x0 of central fringe is obtained by putting n = 0 in equation (1). Therefore,

∴ x0 = − λD

4d

The negative sign shows that the central fringe is obtained at a point O' below the (central) point O.

7. i. a. Two independent monochromatic sources of light cannot produce sustained interference because:
i. If the sources are not coherent, they cannot emit waves continuously.
ii. Independent sources, emit the waves, which don't have the same phase or a constant phase difference. Therefore these
sources will not be coherent and therefore would not produce a sustained interference pattern.
b. given y1​= acosω t,
y2​=a cos (ω t + ϕ),
by superposition principle, resultant displacement, y = y1 ​+ y2,
or y = a cos ω t + a cos(ω t + ϕ)
ϕ ϕ
or y = 2a cos( ).cos(ω t + 2 2
),
ϕ
or y = A cos(ω t + 2
),
it is an equation of simple harmonic plane progressive wave, whose amplitude is A,
ϕ
here A = 2acos( ), 2

now intensity is proportional to square of amplitude , therefore

I = KA2 = 4Ka2cos2( ),
ϕ

where K is proportionality constant.

ii. A path difference of λ , corresponds to a phase difference of 2π
∴ The intensity, K = 4a2 ⇒ a 2
=
K

A path difference of λ

3
, corresponds to a phase difference of 2π

3
ϕ
∴ Intensity = 4a 2
cos
2

2
2π/3
=4×a 2
× cos
2

K 1 2 K
=4× 4
× (
2
) =
4

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8. If coherent sources are not taken, the phase difference between two interfering waves, will change continuously and a sustained
interference pattern will not be obtained. Thus, coherent sources provide a sustained interference pattern.
Let S1 and S2 be two coherent sources separated by a distance d.
Let the distance of the screen from the coherent sources be D. Let point M be the foot of the perpendicular drawn from 0, which is
the midpoint of S1 and S2 on the screen. Obviously, point M is equidistant from S1 and S2. Therefore the path difference between
the two waves at point M is zero. Thus the point M has the maximum intensity.
Now, Consider a point P on the screen at a distance y from M. Draw S1N perpendicular from S1 on S2P.

The path difference between two waves reaching at P from S1 and S2 is given by,
△ = S2P - S1P ≈ S2N
Since, D >> d, so ∠ S2S1N = θ is very small
∠ S2S1N = ∠ MOP = θ
In △S1S2N,
S2 N
sin θ = S1 S2

In △MOP,
MP
tan θ = OM

Since, θ is very small,

sin θ = θ = tan θ
S2 N MP
=
S1 S2 OM

∴ S2N = S1S2 × MP

OM
= d ⋅
Y

Therefore
yd
Path difference, △S2P - S1P = S2N = D
...(1)
For bright fringe or constructive interference, we have
yd

D
= nλ , n = 0, 1, 2, 3 ...
Therefore
nDλ
yn =
d

For dark fringe, path difference is an odd multiple or half wavelength. So, we have
y⋅d

D
= (2n - 1) λ

2
; n = 1, 2, 3, ...
1 Dλ
yn = (n − )
2 d

Expression for Fringe Width:

Distance between any two consecutive bright fringes or any two consecutive dark fringes are called the fringe width.
It is denoted by ω .
Let, yn+1 and yn be the distance of two consecutive fringes. Then, we have
yn+1 = (n + 1) Dλ

d
and yn = nDλ

So, fringe width = yn+1 - yn = (n + 1) Dλ

d
−
nDλ

d
=
Dλ

Fringe width is same for both bright and dark fringe.

When the entire apparatus is immersed in water, the fringe width decreases.
9. Angular width is given by
θ= or d =
λ

d
λ

i. According to the question, λ = 600nm = 6 × 10 −7

m
0.1π π
θ= rad = rad
180 1800
λ
d =
θ
−7
6× 10 ×1800 −4
∴ d = = 3.44 × 10 m
π

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 ii. The frequency of a light depends on its source only.
So, the frequencies of reflected and refracted light will be same as that of incident light.
Reflected light is in the same medium (air).
o

So its wavelength remains same as 500A .

We know that ν = c

λ
8
3×10
=
−10
5000×10

18
= 6 × 10 Hz

This is the required frequency of both refracted and reflected light.

We know that,
speed of light in air
μ=
speed of light in water
8
4 3×10
=
3 v

8
v = 2.25 × 10 m/s

speed of light in water = 2.25 × 10 m/s

8

Wavelength of refracted light is given by λ ′

=
v

ν
= 0.375 × 10
−6
m

So, wavelength of refracted wave will be decreased.

10. a. Law of Reflection: Let XY be a reflecting surface at which a wavefront is being incident obliquely. Let v be the speed of the
wavefront and at time t = 0, the wavefront touches the surface XY at A. After time t, the point B of wavefront reaches the
point B' of the surface. According to Huygens principle, each point of the wavefront acts as a source of secondary waves.
When point A of the wavefront strikes the reflecting surface, then due to the presence of the reflecting surface, it cannot
advance further; but the secondary wavelet originating from point A begins to spread in all directions in the first medium with
speed v. As the wavefront AB advances further, its points A1, A2, A3 K etc. strike the reflecting surface successively and send
spherical secondary wavelets in the first medium.

First of all the secondary wavelet starts from point A and traverses distance AA' (=vt) in the first medium in time t. In the
same time t, the point B of wavefront, after travelling a distance BB', reaches point B' (of the surface), from where the
secondary wavelet now starts. Now taking A as centre we draw a spherical arc of radius AA' (= vt) and draw tangent A' B' on
this arc from point B'. As the incident wavefront AB advances, the secondary wavelets start from points between A and B¢,
one after the other and will touch A' B' simultaneously. According to Huygens principle wavefront A' B' represents the new
position of AB, i.e., A' B' is the reflected wavefront corresponding to incident wavefront AB. Now in right-angled triangles
ABB' and AA' B'
∠ ABB' = ∠ AA'B' (both are equal to 90o)
side BB' = side AA' (both are equal to vt) and side AB' is common i.e., both triangles are congruent.
∴ ∠ BAB' = ∠ AB'A
i.e., incident wavefront AB and reflected wavefront A' B' make equal angles with the reflecting surface XY. As the rays are
always normal to the wavefront, therefore the incident and the reflected rays make equal angles with the normal drawn on the
surface XY, i.e., angle of incidence i = angle of reflection r
This is the second law of reflection. Since AB, A' B' and XY are all in the plane of paper, therefore the perpendiculars dropped
on them will also be in the same plane. Therefore we conclude that the incident ray, reflected ray and the normal at the point
of incidence, all lie in the same plane. This is the first law of reflection. Thus Huygens principle explains both the laws of
reflection.
b. i. If the radiation of a certain frequency interacts with the atoms/molecules of the matter, they start to vibrate with the same
frequency under forced oscillations. Thus, the frequency of the scattered light (Under reflection and refraction) equals to
the frequency of incident radiation.
ii. No, the energy carried by the wave depends on the amplitude of the wave, but not on the speed of the wave.
Section B

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11. i. According to Faraday's flux rule, the magnitude of induced emf is equal to the rate of change of magnetic flux linked with the
closed circuit.
dϕ
ε= −
dt

ii. By faraday's laws of electromagnetic induction,

dϕ
V =
dt
dA
V = B
dt

V = Blv as dA

dt
= lv
This is the induced emf across the ends of the rod.
→ −
→ →
iii. We have the force on the free electrons from Q to P as F b = qv × B .
where v is the velocity of the rod as well as of the free electrons inside it, B is the uniform magnetic field. The free electrons
will move towards P and the positive charge will appear at Q. An electrostatic field is developed from Q to P in the wire
→ →
which exerts a force F e = qE on each electrons. The charge keeps on gathering until.
→ →
Fb = Fe and the resultant force on each electron is zero.
→ →
→
|q v × B | = |q E |

⇒ vB = E
The potential difference between Q and P is then
V = E = vBl
which is maintained by the magnetic force of the moving electron producing an emf, e = Bvl

12. i. The deflection is caused by a phenomenon called electromagnetic induction. The amount of deflection in the galvanometer
depends upon the speed with which the magnet is moved. The direction of deflection in the galvanometer depends on the
direction (towards/away) of the movement of the magnet. The laws of electromagnetic induction are:
a. Whenever the magnetic flux linked with a closed circuit changes, an induced emf is produced in it.
b. The magnitude of the induced emf is equal to the rate of change of magnetic flux linked with the closed circuit.
dϕ
ε= −
dt

ii. The flux enclosed by the rod is

Φ = Blx for 0 ≤ x ≤ b

= Blx for b ≤ x ≤ 2b
Now, the magnitude of induced emf acording to faraday's law is
dϕ
ε=
dt
= −Bl
dx

dt
= −Blv for 0 ≤ x ≤ b
= −Bl
db

dt
= 0 for b ≤ x ≤ 2b
Now, the magnitude of induced current, when induced emf is non-zero is ε = I R
ε Blv
∴ I = =
r r

The force required to keep the conductor in motion is F = Bll

2 2
Blv B l v
∴ F = B l =
R R
2 2

∴ F =
B l v

R
for b ≤ x < b
= 0 for b ≤ x ≤ 2b
Therefore, the variation of flux, emf, and force is as shown in the following diagram

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13. Consider a conducting coil of N turns and area S rotating with a uniform angular velocity ω about an axis in the plane of the coil,
and perpendicular to a uniform magnetic field of induction B.

The frequency of rotation of the coil is, f = ω

2π
Magnetic flux through the loop in the given position is given by:
ϕ = ⃗
B ⋅ S
⃗

= BS cos ( π

2
− θ)

= BS sin (θ)
We have, θ = θ0 + ω t
Therefore, ϕ = BS sin (θ0 + ω )
Induced emf is given by,
dϕ
ε = -N dt

Pitting ϕ = BS sin (θ0 + ω t)

d(ϕ=BS sin( θo +ωt))
ε = -N dt

ε = -NBS ω cos (θ0 + ω t)

Therefore, the e.m.f induced in coil rotating in uniform magnetic field is given by ε = −NBSω cos(θ0 + ω t).
Where θ is angle made by area vector with the magnetic field at any particular instant of time.
When the coil is rotated from 0o, the emf keeps increasing until the coil is parallel to the field and then keeps decreasing until it
turns 180o. Then on further rotating the value of emf increases but the direction is reversed until it is again parallel to the field. On
further rotating to 360o it reaches zero.
The instantaneous value of induced emf,

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 E = E0 sin (90o − 60o)

= E0 sin 30o =
E0

14. a.

i. ε ′
= Blv
v
ε = Bl
2

lω
ε = Bl ( )
2

2
bl ω
ε=
2
2
ε bl ω
I = =
R 2R

ii. F = I ( l × b)
3
bl ωB
F =
2R

Direction of force is perpendicular to both l ⃗ and B⃗

2 2
bl ω
iii. Power = i 2
R= (
2R
) R

2 4 2
b l ω
=
4R

b. Since induced current will reduce, it will be a little easier to remove the coil
15. a. i. Whenever there is a change in magnetic flux linked with a coil, an emf is induced in the coil, however it lasts so long as
magnetic flux keeps on changing.
ii. The magnitude of the induced emf is equal to the rate of change of magnetic flux through the circuit.
b.

Straight conductor PQ of length ‘l’ is moving with velocity ‘v’ in uniform magnetic field B, which is perpendicular to the
plane of the system.
Length RQ = x, RS = PQ = l
Instantaneous flux = (normal) field × area
The magnetic flux (ϕ B) enclosed by the loop PQRS,
∴ ϕB = Blx

Since ‘x’ is changing with time, there is a change of magnetic flux. The rate of change of this magnetic flux determines the
induced emf
−dϕ −d
∴ e= = (Blx)
dt dt
dx
= −Bl
dt

e = Blv
dx
as dt
= −v

c. Work done (that gets stored as the magnetic potential energy) when current ’I’ flows in the solenoid.
dW = (e) (Idt)
dI
∴ dW = (L ) . (I dt)
dt

∴ dW = LIdI

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Total work done W = ∫ dW = ∫ LI dI
1 2
W = LI
2

For the solenoid, we have L = μ 2

0 n Al and B = μ
0 nI

2
1 2 B
∴ W = (μ0 n Al) [ ]
2 μ n
0

2
B Al
=
2μ0

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